If you add the denominator to both the numerator and denominator of a fraction, you get a new fraction that sometimes is larger than the original fraction.
Find a fraction where the new fraction is four times as large as the original fraction after you add the denominator to both the numerator and denominator.
Extra: In order for the new fraction to be n times greater than the original fraction, how many times larger than the original numerator must the original denominator be? Express your answer in terms of n.
Please help me its going in my grade!!!
(n+d)/(d+d) = 4n/d
nd+d^2 = 8nd
d^2-7nd = 0
d-7n = 0
d = 7n
so, you can have
1/7: 8/14 = 4/7
3/21: 24/42 = 4/7
and so on
Now, replace 4 with k and see what you get to have the new fraction k times as large.
original fraction -- n/d
new fraction formed = (n+d)/2d
you want: (n+d)/2d = 4(n/d)
8nd = nd + d^2
7nd = d^2
7n = d
it looks like we can find an infinite number of cases like that.
e.g. let n = 1, then d = 7
then the original fraction was 1/7 and
the fraction formed by your rule = 8/14, which is 4/7 and it is 4 times as large
let n = 13 , then the original fraction is
13/91 and the new fraction is 104/182
now 13/91 = 1/7, and 104/182 = 4/7
etc
notice it works even for fractions not in lowest terms, after your reduce them they all reduce to 1/7 and 4/7
To find a fraction where the new fraction is four times as large as the original fraction after adding the denominator to both the numerator and denominator, we can proceed step by step:
Let the original fraction be represented as a/b, where a is the numerator and b is the denominator.
Step 1: Add the denominator to both the numerator and denominator of the original fraction:
(a + b) / (b + b)
Step 2: Simplify the expression of the new fraction:
(a + b) / (2b)
Step 3: Set up the equation for the new fraction being four times as large as the original fraction:
4(a/b) = (a + b) / (2b)
Step 4: Cross-multiply to solve for a:
4(a/b) = (a + b) / (2b)
4a = (a + b) / 2
8a = a + b
7a = b
So, for the new fraction to be four times as large as the original fraction, the original numerator (a) must be equal to one-seventh of the original denominator (b).
Extra: If we want the new fraction to be n times greater than the original fraction, the original numerator (a) would need to be equal to 1/n of the original denominator (b).
To find a fraction where the new fraction is four times as large as the original fraction, we can set up an equation:
Let the original fraction be represented as a/b, where 'a' is the numerator and 'b' is the denominator.
After adding the denominator to both the numerator and denominator, the new fraction becomes (a+b)/(b+b).
We want this new fraction to be four times as large as the original fraction, so we can write the equation as:
4 * (a/b) = (a+b)/(2b)
To solve for the original fraction, we can cross-multiply:
4a * 2b = (a+b) * b
8ab = ab + b^2
Rearranging the equation, we get:
b^2 - 7ab = 0
Now, we need to find a fraction where the new fraction is n times greater than the original fraction after adding the denominator to both the numerator and denominator. We'll represent the original fraction as a/b, where 'a' is the numerator and 'b' is the denominator.
Following similar steps as before, we set up the equation:
n * (a/b) = (a+b)/(b+b)
Cross-multiplying, we have:
na * 2b = (a+b) * b
2nab = ab + b^2
Rearranging, we get:
b^2 - (2na)b = 0
So, the original numerator must be (2na) times larger than the original denominator in order for the new fraction to be n times greater than the original fraction.